KNEC Electrical Power Systems and Electromagnetic Field Theory Past Paper with Answers — June/July 2025

KNEC Diploma in Electrical and Electronic Engineering (Power Option), Module III
Paper codes: 2521/305 and 2601/305 · Session: June/July 2025 · Time: 3 hours

Revision note: Attempt the questions before reading the solutions. These independently prepared answers are for revision and are not an official KNEC marking scheme. The paper contains eight 20-mark questions; answer any three from Section A and any two from Section B.

SECTION A: ELECTRICAL POWER SYSTEMS

Question 1 (20 marks)

(a)(i) State three factors which affect conductor vibrations.

Answer (a)(i). Three important factors are:

1. Wind speed and turbulence: Wind produces the aerodynamic forces that excite the conductor. Steady crosswind can cause periodic vortex shedding; turbulent or stronger wind can increase vibration amplitude.
2. Conductor properties: Diameter, mass per unit length, surface stranding and mechanical damping affect the natural frequency and response to wind.
3. Span and tension: Span length and conductor tension determine natural modes and vibration amplitude. Ice or an uneven surface coating can change the aerodynamic response.

(a)(ii) Explain high-frequency vibration and galloping.

Answer (a)(ii). High-frequency (Aeolian) vibration is a small-amplitude, relatively high-frequency oscillation, commonly caused by alternating vortices shed from a conductor in a light-to-moderate steady wind. Repeated bending near clamps can cause fatigue damage; Stockbridge dampers reduce it. Galloping is a low-frequency, large-amplitude movement, often produced when wind acts on an iced or asymmetrically coated conductor. The conductor may move vertically or elliptically, creating clearance problems and possible phase clashing.

(b) Explain how atmosphere and conductor size affect corona.

Answer (b). Atmosphere: Reduced air density at high altitude, high temperature or low pressure lowers the critical disruptive voltage. Rain, dust and other surface contamination can increase local electric-field concentration, corona loss, noise and radio interference. Conductor size: A larger conductor radius reduces the electric-field gradient at its surface. It raises the disruptive voltage and therefore reduces corona at a given operating voltage.

(c) A 400 kV, 50 Hz, three-phase overhead line has conductor radius 2 cm and equilateral spacing 3 m. Ambient temperature is 50 °C, atmospheric pressure is 76 cm Hg and the irregularity factor is 0.85. Determine (i) air-density factor, (ii) critical disruptive voltage and (iii) total corona loss for the three phases.

Answer (c). Convert the spacing to 300 cm. The air-density factor is:

δ = 3.92b/(273 + t) = (3.92 × 76)/(273 + 50) = 0.92235

For a three-phase line, the phase voltage is:

V_ph = 400/√3 = 230.94 kV rms

Peek’s disruptive-voltage formula uses r and D in centimetres:

V_d = 21.1 m₀ δ r ln(D/r) = 21.1 × 0.85 × 0.92235 × 2 × ln(300/2) = 165.78 kV per phase

Since V_ph > V_d, use Peek’s fair-weather corona-loss formula. It gives loss per phase per kilometre:

P_c,phase = 242.2(f + 25)/δ × √(r/D) × (V_ph − V_d)² × 10⁻⁵ = 68.28 kW/km

P_c,3-phase = 3 × 68.28 = 204.85 kW/km

Answers: δ = 0.92235; V_d = 165.78 kV per phase; total corona loss = 204.85 kW per kilometre for all three phases. The paper gives no line length, so an absolute whole-line loss cannot be calculated.

Question 2 (20 marks)

(a) State two factors which affect conductor sag.

Answer (a). Sag increases with conductor weight and span length, and decreases as horizontal tension increases. Temperature also affects sag through thermal expansion.

(b) An overhead-transmission line crossing a river is supported by two towers 40 m and 100 m above water level, with a 500 m span. The conductor weighs 2 kg/m and the working tension is 1,400 kg. Determine (i) the conductor distance at the lower-most support and (ii) the minimum clearance from the water.

Answer (b). Treat the given kilogram units as kgf and use the parabolic sag approximation. Let x be the horizontal distance from the lower support to the conductor’s lowest point. For support-height difference h = 60 m:

x = L/2 − Th/(wL) = 500/2 − (1,400 × 60)/(2 × 500) = 166 m

The vertical sag below the lower support is:

f_lower = wx²/(2T) = 2 × 166²/(2 × 1,400) = 19.68 m

Minimum water clearance = 40 − 19.68 = 20.32 m

Answers: the lowest point is 166 m horizontally from the lower support and the sag there is 19.68 m below that support; minimum clearance above water is 20.32 m. This states both distances to resolve the paper’s wording in part (i).

(c)(i) State two merits of the symmetrical-components method in solving unsymmetrical faults.

(c)(ii) A three-phase, 20 MVA, 22 kV generator with an earthed neutral supplies a feeder. Generator and feeder sequence impedances are listed below in ohms. For a single line-to-ground fault on the red phase, determine (I) total impedance and (II) total fault current.

Answer (c)(i). The method converts an unbalanced three-phase system into positive-, negative- and zero-sequence networks that are easier to analyse. It gives a systematic way to connect the sequence networks for each fault type and calculate phase currents and voltages.

Answer (c)(ii). The printed sequence impedances are:

Sequence	Generator	Feeder	Total
Positive	j1.5 Ω	j1.2 Ω	j2.7 Ω
Negative	j0.8 Ω	j1.2 Ω	j2.0 Ω
Zero	j0.6 Ω	j4.0 Ω	j4.6 Ω

For a bolted single line-to-ground fault with a directly earthed neutral, the sequence networks are in series:

Positive sequence       Negative sequence       Zero sequence
E₁ ── Z₁ ─────────────── Z₂ ─────────────────── Z₀ ── ground
              I₁ = I₂ = I₀

Z_total = Z₁ + Z₂ + Z₀ = j(2.7 + 2.0 + 4.6) = j9.3 Ω

Phase voltage and fault current are:

V_ph = 22/√3 = 12.702 kV

I_f = 3V_ph/Z_total = 3 × 12.702 kV/(j9.3 Ω) = 4.097 kA ∠−90°

Answers: Z_total = j9.3 Ω; fault-current magnitude = 4.10 kA. No neutral impedance is specified, so the neutral is treated as solidly earthed.

Question 3 (20 marks)

(a) Explain the effect of load power factor on transmission-line efficiency.

Answer (a). For a fixed real load power and voltage, a lower power factor requires more current: I = P/(√3 V_L pf). The higher current increases I²R loss and heating, reducing efficiency. Improving a lagging power factor reduces current and copper loss, so a larger fraction of the sending-end power reaches the load.

(b) With circuit and phasor diagrams for a lagging power factor, show the expression for the sending-end voltage magnitude of a short transmission line.

Answer (b). A short line is modelled by its series resistance and inductive reactance, with shunt capacitance neglected:

Per-phase circuit:      I →
Vₛ ───────── R ───── jXₗ ───── Vᵣ ── load

Phasors, with I as reference:
                         jIXₗ
                           ↑
Vᵣ at +φ  ↗       IR →    │
I reference ─────────────────────────→

Taking the current as the horizontal reference, the receiving-end voltage leads it by φ_R for a lagging load. The in-phase component of V_S is V_R cos φ_R + IR; its quadrature component is V_R sin φ_R + IX_L. Therefore:

V_S = (V_R cos φ_R + IR) + j(V_R sin φ_R + IX_L)

|V_S| = √[(V_R cos φ_R + IR)² + (V_R sin φ_R + IX_L)²]

(c) A 50 Hz, three-phase overhead line is 100 km long and supplies 50 MW at 66 kV, 0.8 power factor lagging. Per phase per kilometre, R = 0.2 Ω, X_L = 0.4 Ω and capacitive susceptance is 5 × 10⁻⁶ S. Using the nominal-π method, determine (i) receiving-end current, (ii) line current and (iii) sending-end voltage.

Answer (c). Work per phase. Total series impedance and shunt admittance are:

Z = 100(0.2 + j0.4) = 20 + j40 Ω; Y = j100(5 × 10⁻⁶) = j0.0005 S

The receiving-end phase voltage and load current are:

V_R = 66/√3 = 38.105 kV

I_R = 50 × 10⁶/(√3 × 66 × 10³ × 0.8) ∠−cos⁻¹(0.8) = 546.73∠−36.87° A

Receiving-end shunt current, then current through the series branch:

I_cR = (Y/2)V_R = j9.526 A

I_line = I_R + I_cR = 437.387 − j318.514 = 541.07∠−36.06° A

Hence the sending-end phase voltage is:

V_S = V_R + ZI_line = 59.593 + j11.125 = 60.623∠10.57° kV per phase

|V_S,line| = √3 × 60.623 = 105.00 kV line-to-line

Answers: receiving-end load current = 546.73∠−36.87° A; series line current = 541.07∠−36.06° A; sending-end voltage = 105.00 kV line-to-line. If “line current” means the current at the sending terminals, adding the sending-end shunt current gives I_S = 530.15∠−34.94° A.

Question 4 (20 marks)

(a) Explain the effect of arcing-ground and open-circuited-line surges on a transmission line.

Answer (a). Arcing ground: An intermittent arc from a phase to earth repeatedly extinguishes and restrikes. The resulting oscillatory transients can produce high overvoltages on the healthy phases and stress line insulation. Open-circuited line: At an open end, the voltage reflection coefficient is +1, so the reflected voltage has the same polarity as the incident surge. The end voltage can momentarily approach twice the incident voltage, risking insulation flashover.

(b) A 200 kV voltage surge travels on an overhead line connected in series with an underground cable. Per kilometre, the overhead-line inductance and capacitance are 2 mH and 0.095 μF; the cable values are 0.3 mH and 0.4 μF. Determine (i) the surge impedance of the line and cable and (ii) the reflected and transmitted surge voltages at the cable entrance.

Answer (b). For a lossless line, Z₀ = √(L/C):

Z_overhead = √[(2 × 10⁻³)/(0.095 × 10⁻⁶)] = 145.10 Ω

Z_cable = √[(0.3 × 10⁻³)/(0.4 × 10⁻⁶)] = 27.39 Ω

For an incident surge travelling from the overhead line into the cable:

Γ_V = (Z_cable − Z_overhead)/(Z_cable + Z_overhead) = −0.68245

V_reflected = Γ_VV_incident = −136.49 kV; V_transmitted = (1 + Γ_V)V_incident = 63.51 kV

Answers: overhead-line surge impedance = 145.10 Ω; cable surge impedance = 27.39 Ω; reflected wave = −136.49 kV toward the source; transmitted wave = 63.51 kV into the cable.

(c)(i) State two reasons for busbar protection.

(c)(ii) A star-connected, three-phase, 20 MVA, 6.6 kV alternator is protected by the Merz-Price circulating-current principle using 1000/10 A current transformers. The star point is earthed through 10 Ω. The minimum relay operating current is 0.5 A. Determine (I) minimum fault current and (II) the percentage of each stator winding unprotected against an earth fault.

Answer (c)(i). Busbars are critical junctions; a fault can interrupt several circuits and produce a very high fault current. Fast, selective busbar protection limits equipment damage and avoids unnecessarily disconnecting healthy circuits.

Answer (c)(ii). The CT ratio is 1000/10 = 100:1, so the relay pickup corresponds to this primary fault current:

I_f,min = 0.5 × 100 = 50 A

For a fault a fraction x of the phase winding from the neutral, the voltage across the neutral resistor is xV_ph. At the protection boundary, xV_ph/R_n = I_f,min:

V_ph = 6.6/√3 = 3.8105 kV

x = 500/3,810.5 = 0.1312 = 13.12%

Answers: minimum primary fault current = 50 A; approximately 13.12% of each phase winding nearest the neutral is unprotected (about 86.88% is protected).

Question 5 (20 marks)

(a) Define steady-state stability and dynamic stability in power systems.

Answer (a). Steady-state stability is the ability of a power system to remain in synchronism after a small, gradual change in load or operating condition. Dynamic stability is the ability to maintain synchronism after small disturbances while the system’s automatic controls and damping responses are included over time.

(b) A 50 Hz synchronous generator has internal voltage 1.4 p.u. and reactance 0.5 p.u. It is connected to an infinite bus through a double-circuit line; each circuit has reactance 0.35 p.u. The generator delivers 0.8 p.u. power to a 1.0 p.u. infinite-bus voltage. Sketch the line diagram and determine (i) maximum power and (ii) steady-state operating angle.

Answer (b). The two identical line circuits are in parallel:

Generator internal emf              Infinite bus
E′ = 1.4 p.u. ── j0.5 p.u. ──●──┬── j0.35 p.u. ──┬── V = 1∠0 p.u.
                             └──┴── j0.35 p.u. ──┘
                                  double circuit

X_line,eq = 0.35 ∥ 0.35 = 0.175 p.u.; X_total = 0.5 + 0.175 = 0.675 p.u.

The power-angle relation (neglecting resistance) is P = EV/X sin δ:

P_max = EV/X_total = (1.4 × 1.0)/0.675 = 2.074 p.u.

0.8 = 2.074 sin δ ⇒ δ = sin⁻¹(0.8/2.074) = 22.69°

Answers: maximum power = 2.074 p.u.; stable operating angle = 22.69°.

(c) With the aid of a diagram, describe the operation of a Buchholz transformer-protection system.

Answer (c). A Buchholz relay is fitted in the oil pipe between the main tank and conservator of an oil-immersed transformer. It responds to gas formation and oil movement caused by internal faults:

Conservator
    │
    ├── oil pipe ── [ Buchholz relay ] ── Main transformer tank
                    upper float → alarm
                    lower vane/float → trip circuit breaker
                    gas-collection pocket → gas test/inspection

A minor or incipient internal fault heats and decomposes oil, producing gas that rises and collects in the relay. The oil level falls, tilting the upper float and closing the alarm contacts. A severe internal fault produces a rapid gas generation and oil surge toward the conservator; the lower vane/float operates the trip contacts and opens the transformer circuit breaker. The collected gas can be examined to help diagnose the fault. This protection applies to oil-filled transformers with a conservator and does not replace protection for faults outside the tank.

SECTION B: ELECTROMAGNETIC FIELD THEORY

Question 6 (20 marks)

(a) State two sources and two applications of electromagnetic-wave radiation.

Answer (a). Sources: accelerating electric charges, such as alternating currents in transmitting antennas, and natural sources such as the Sun or other hot luminous bodies. Applications: radio/television and mobile communications; radar or medical imaging and heating, depending on the waveband.

(b) The electric-flux density is D = (2y² + z)a_x + 4xy a_y + x a_z C/m². Determine (i) volume charge density at (−1, 0, 3) and (ii) flux through the cube 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1.

Answer (b). By Gauss’s law, ρ_v = ∇·D:

ρ_v = ∂(2y² + z)/∂x + ∂(4xy)/∂y + ∂x/∂z = 4x C/m³

ρ_v(−1, 0, 3) = −4 C/m³

The outward flux through the closed cube follows from the divergence theorem:

Ψ = ∭_V(∇·D)dV = ∫₀¹∫₀¹∫₀¹ 4x dz dy dx = 2 C

Answers: volume charge density = −4 C/m³; outward electric flux = 2 C.

(c)(i) State the Biot–Savart law, giving its expression.

(c)(ii) In a conducting region, H = [yz(x² + y²)a_x − y²xz a_y + 4x²y² a_z] A/m. Determine the current density at (5, 2, −3).

Answer (c)(i). The Biot–Savart law gives the magnetic-field contribution from a steady current element. Its differential form is:

dH = I(dℓ × a_R)/(4πR²)

Integrating over the current path gives the total H field.

Answer (c)(ii). In a steady conducting region, J = ∇×H. With H_x = yz(x² + y²), H_y = −y²xz and H_z = 4x²y²:

J_x = ∂H_z/∂y − ∂H_y/∂z = 8x²y + xy²

J_y = ∂H_x/∂z − ∂H_z/∂x = y(x² + y²) − 8xy²

J_z = ∂H_y/∂x − ∂H_x/∂y = −z(x² + 4y²)

J(5, 2, −3) = 420a_x − 102a_y + 123a_z A/m²

Question 7 (20 marks)

(a) Write three of Maxwell’s equations in differential form and cite their significance in electromagnetics.

Answer (a).

∇·D = ρ_v — electric charge is the source of electric flux density.

∇×E = −∂B/∂t — a time-varying magnetic field induces a circulating electric field.

∇×H = J + ∂D/∂t — conduction current and changing electric flux produce a magnetic field.

(b) With reference to electromagnetic-wave properties, define (i) skin depth and (ii) lossy dielectric.

Answer (b). Skin depth is the distance into a conducting material at which the field amplitude has fallen to 1/e (about 36.8%) of its surface value; for a good conductor, δ_s ≈ √[2/(ωμσ)] = 1/α. A lossy dielectric has finite conductivity or dielectric loss, so part of an electromagnetic wave’s energy is dissipated as heat while the wave propagates. Its loss tangent for conduction loss is tan δ = σ/(ωε).

(c) In a charge-free region, σ = 0, ε = ε₀ε_r and μ = μ₀. The magnetic-field intensity is H = 5 cos(10⁹t − 4y)a_z A/m. Determine (i) displacement-current density J_d, (ii) magnetic-flux density and (iii) relative permittivity ε_r.

Answer (c). The printed paper labels the magnetic-flux density as D; D normally denotes electric-flux density, while magnetic-flux density is B. The phase constant is β = 4 rad/m and angular frequency is ω = 10⁹ rad/s. Ampère–Maxwell law gives J_d = ∇×H because σ = 0:

J_d = ∇×H = 20 sin(10⁹t − 4y)a_x A/m²

Using μ = μ₀, the requested magnetic-flux density is:

B = μ₀H = 6.283 × 10⁻⁶ cos(10⁹t − 4y)a_z T

If the examiner intended D, the electric-flux density is also found from ∇×H = ∂D/∂t. The wave travels in +y; with H along +z, D (and E) points in −x:

D = −2.00 × 10⁻⁸ cos(10⁹t − 4y)a_x C/m²

For a uniform lossless wave, β = ω√(μ₀ε₀ε_r). Therefore:

ε_r = (βc/ω)² = (4 × 3.00 × 10⁸/10⁹)² = 1.44

Answers: J_d = 20 sin(10⁹t − 4y)a_x A/m²; B = 6.283 μT cos(10⁹t − 4y)a_z; ε_r ≈ 1.44. The additional D expression resolves the notation inconsistency in the scan.

Question 8 (20 marks)

(a) With reference to energy and momentum in electromagnetics, state (i) Poynting’s theorem in integral form and (ii) two properties of electromagnetic waves.

Answer (a)(i). For a linear, isotropic medium, let S = E×H and u = ½(E·D + B·H). Poynting’s theorem is:

∯_SS·da = −d/dt ∭_Vu dV − ∭_VJ·E dV

The net outward electromagnetic power equals the decrease in stored field energy minus the power delivered to charges in the volume.

Answer (a)(ii). Electromagnetic waves are transverse: E and H are perpendicular to each other and to the direction of propagation. They transport energy and momentum in the direction of the Poynting vector S = E×H.

(b) Using energy conservation, explain energy flow in (i) a resistive wire and (ii) a battery source.

Answer (b). Resistive wire: The electromagnetic field carries energy through the space around the wire and the Poynting vector points inward through the wire’s surface. The inward power is converted to heat at the rate J·E, giving I²R loss. Battery: Chemical reactions do work on charges inside the source, converting chemical energy into electrical field energy. The battery supplies electromagnetic energy to the surrounding fields and circuit at approximately VI; that energy flows to the load, where it is converted to heat or useful work.

(c) A plane wave in a non-magnetic medium has E = 60 sin(10¹⁰t + 2z)a_y V/m. Determine (i) direction of propagation, (ii) wavelength λ, frequency f, relative permittivity ε_r and intrinsic impedance η (printed as R).

Answer (c). The phase is ωt + βz, so constant phase moves in the −z direction. Here ω = 10¹⁰ rad/s and β = 2 rad/m:

Direction of propagation = −a_z

λ = 2π/β = π = 3.142 m

f = ω/(2π) = 10¹⁰/(2π) = 1.592 × 10⁹ Hz = 1.592 GHz

With μ_r = 1 and c ≈ 3 × 10⁸ m/s:

v_p = ω/β = 5.00 × 10⁹ m/s; ε_r = (βc/ω)² = 0.0036

η = η₀/√ε_r = 376.73/√0.0036 = 6.279 × 10³ Ω

Answers: wave travels in −z; λ = 3.142 m; f = 1.592 GHz; ε_r = 0.0036; intrinsic impedance η ≈ 6.28 kΩ. The printed ω and β imply a phase velocity above c and an unusually low ε_r; the calculations follow the printed values, which may contain a source typo.